题目描述

题解：

裸如飞行员的二分图匹配问题。

直接上代码：

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define K 50
#define N 1250
const int inf = 0x3f3f3f3f;
inline int rd()
{
    int f=1,c=0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=10*c+ch-'0';ch=getchar();}
    return f*c;
}
int k,n,m,S,T,hed[N],cnt=-1,cur[N];
struct EG
{
    int to,nxt,w;
}e[N*K*2];
void ae(int f,int t,int w)
{
    e[++cnt].to = t;
    e[cnt].nxt = hed[f];
    e[cnt].w = w;
    hed[f] = cnt;
}
int dep[N];
bool vis[N];
queue<int>q;
bool bfs()
{
    memset(dep,0x3f,sizeof(dep));
    memcpy(cur,hed,sizeof(cur));
    dep[S]=0,vis[S]=1;q.push(S);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int j=hed[u];~j;j=e[j].nxt)
        {
            int to = e[j].to;
            if(e[j].w&&dep[to]>dep[u]+1)
            {
                dep[to] = dep[u]+1;
                if(!vis[to])
                {
                    vis[to] = 1;
                    q.push(to);
                }
            }
        }
        vis[u] = 0;
    }
    return dep[T]!=inf;
}
int dfs(int u,int lim)
{
    if(u==T||!lim)return lim;
    int fl = 0,f;
    for(int j=cur[u];~j;j=e[j].nxt)
    {
        cur[u] = j;
        int to = e[j].to;
        if(dep[to]==dep[u]+1&&(f=dfs(to,min(lim,e[j].w))))
        {
            fl+=f,lim-=f;
            e[j].w-=f,e[j^1].w+=f;
            if(!lim)break;
        }
    }
    return fl;
}
int dinic()
{
    int ret = 0;
    while(bfs())ret+=dfs(S,inf);
    return ret;
}
int main()
{
    k = rd(),n = rd();
    S = n+k+1,T=n+k+2;
    memset(hed,-1,sizeof(hed));
    for(int x,i=1;i<=k;i++)
    {
        x = rd();
        ae(S,i,x);
        ae(i,S,0);
        m+=x;
    }
    for(int c,x,i=1;i<=n;i++)
    {
        c = rd();
        ae(i+k,T,1);
        ae(T,i+k,0);
        while(c--)
        {
            x = rd();
            ae(x,i+k,1);
            ae(i+k,x,0);
        }
    }
    int tmp = dinic();
    if(tmp==m)
    {
        for(int i=1;i<=k;i++)
        {
            printf("%d: ",i);
            for(int j=hed[i];~j;j=e[j].nxt)
            {
                int to = e[j].to;
                if(to==S)continue;
                if(e[j].w)continue;
                printf("%d ",to-k);
            }
            puts("");
        }
    }else
    {
        puts("No Solution!");
    }
    return 0;
}